Optimisation : Maxima & Minima

Problem Solving

Differentiation can be used to solve problems which require

maximum or minimum values.

Problems typically cover topics such as areas, volumes and

rates of change. They often involve having to establish a

suitable formula in one variable and then differentiating to

find a maximum or minimum value. This is known as

Optimisation.

It is important to check the validity of any solutions as often

an answer is either nonexistent (e.g. a negative length or time)

or outside an acceptable interval.

Optimisation

Optimisation is the term we use to describe a simple process

of finding a maximum or minimum value for a given

situation. Normally we would have to graph functions and

then use our graph to establish a maximum or minimum.

As we have learned previously, differentiation allows us to

quickly find the required value and is the expected manner of

solving problems like this in Higher mathematics.

Drawing graphs would not be an acceptable solution.

Optimisation Problems posed will often involve more than one variable.

The process of differentiation requires that we rewrite or re-

arrange formulae so that there is only one variable – typically

x.

Questions are therefore multi-part where the first part would

involve establishing a formula in x from a given situation, part

2 would involve the differentiation and validation of an

acceptable answer, with part 3 the solution to the problem set.

You would do well to note that although the first part is

important, you can normally expect to complete the rest of the

question even when you cannot justify a formula in part 1.

Optimization : Maxima and Minima

•Differentiation is most commonly used to solve problems

by providing a “best fit” solution.

•Maximum and minimum values can be obtained from the

Stationary Points and their nature.

•In exams you may be asked to “prove” a particular formula is valid.

Even if you cannot prove this USE THIS FORMULA TO

ANSWER THE REST OF THE QUESTION.

Copy the following:

Example 1 (Formula Given)

The “Smelly Place” Garden Centre has a model train which is used

to show visitors around the many displays of plants and flowers.

It has been established that the cost per month, C, of running the train

is given by:

40050 25C V

V

where V is the speed in miles per hour.

Calculate the speed which makes the cost per month a minimum and

hence calculate this cost.

Example 1 400

50 25C VV

Solution:

1. Find derivative, remembering

to prepare for differentiation.

150 25 400C V V

225 400dC

VdV

2

40025

dC

dV V

Example 1 400

50 25C VV

Solution:

2. Make statement then

set derivative equal to zero.

2

400 25

1V

2

4000 25

V

2

40025

dC

dV V

At SP’s dC/dV = 0

Cross Multiply

2400 25V2 16V

4V

Reject

–ve speed

Example 1 400

50 25C VV

Solution:

3. Justify nature of each SP

using a nature table.

2

40025

dC

dV V 4V

X

Slope

dC

dV

44 4

-ve 0 +ve

A minimum tp occurs and so cost

is minimised at V = 4. 4. Make statement

2

40025 375

(1) (1)

dC

d

2

40025 17

(5) (5)

dC

d

Example 1 400

50 25C VV

Solution:

4. Sub x-value found into

original expression to find

corresponding minimum value.

A minimum tp occurs and so cost

is minimised at V = 4.

40050 25C V

V

40050 25(4)

4C

For V = 4:

50 100 100C

C = £250

5. Answer Question Minimum cost is £250 per month

and occurs when speed = 4mph

(DO NOT USE dy/dx)

Heinemann , p.115, EX 6R

Q3

Example 2 (Perimeter and area)

A desk is designed which is rectangular in shape and which is required

in the design brief to have a perimeter of 420 cm.

If x is the length of the base of the desk:

(a) Find an expression for the area of the desk, in terms of x.

(b) Find the dimensions which will give the maximum area.

(c) Calculate this maximum area.

Example 2

Solution to (a):

1. Start with what you know

2. In order to differentiate we

need expression only involve

X’s. Use other information to

find an expression for Y. 420 2 2x y

Area = length x breadth

Lets call breadth y, so: Area = x x y

Perimeter = 2x + 2y

3. Change subject to y.

2 420 2y x

210y x

Divide

by 2

x

y

Example 2

Solution to (a):

4. Now substitute this expression

for y into our original expression

for AREA.

( ) 210 2A x x

2( ) 210Area A x x x

At SP’s A’(x) = 0

210 2 0x

2 210x

105x

Area = x x y 210y x

Area = x x (210 – x)

5. To find dimensions giving

maximum or minimum first find

derivative.

6. Make statement and set

derivative equal to zero.

Solution to (b):

Example 2

Solution to (a):

7. Now use your expression for

y to find other dimension.

105x

Area = x x y 210y x

210 105 105y

Example 2

Solution:

8. Must justify maximum

values using nature table.

X

Slope

( )A x

105

+ve 0 -ve

The area is maximised when

length is 105cm and breadth is

105 cm.

9. Make statement

( ) 210 2A x x

X = 105 and Y = 105

(100) 210 2(100) 10A

(110) 210 2(110) 10A

Example 2

Solution to (c):

10. To find actual maximum area

sub maximum x-value just

found into expression for area.

max 22,050 11,025Area (DO NOT USE dy/dx)

The area is maximised when

length is 105cm and breadth is

105 cm.

2( ) 210Area A x x x

2

max 210(105) (105)Area

2

max 11,025Area cm

PROOF

Heinemann , p.112, EX 6Q

Q1, 2, 5

Example 3 (Surface Area and Volume)

A cuboid of volume 51.2 cm3 is made with a length 4 times its

breadth.

If x cm is the breadth of the base of the cuboid:

(a) Find an expression for the surface area of the cuboid, in terms of x.

(b) Find the dimensions which will give the minimum surface area

and calculate this area.

xcm

Example 3

Solution to (a):

1. Start with what you know

2. In order to differentiate

expression must only involve

X’s. Use other information to

find an expression for h.

251.2 4x h

Surface Area = sum of area of faces

Lets call height h, so: = 2(4x x h)

Volume = 4x x x x h

3. Change subject to h. 2 2

51.2 12.8

4h

x x

x 4x

h

+ 2(x x h) + 2(4x x x)

= 10xh + 8x2

4. Now substitute this expression

for y into our original expression

for SURFACE AREA.

Example 2

Solution to (a):

2

2

12.8( ) 10 8A x x x

x

Surface Area = sum of area of faces

2

12.8h

x

2

2

128( ) 8

xA x x

x

2 128( ) 8A x x

x

= 10xh + 8x2

Example 3

Solution to (b):

5. To find dimensions giving

maximum or minimum first find

derivative.

2 128( ) 8A x x

x

2

128( ) 16A x x

x

2 1( ) 8 128A x x x

6. Make a statement and set

derivative equal to zero.

At SP’s A’(x) = 0

2

12816 0x

x

316 128 0x

Multiply

by x2

316 128x

3 8x 3 8 2x

Example 3

Solution to (b):

7. To find corresponding height

substitute x value just found into

expression for h found in step 2. So dimensions minimising area:

2

12.8 12.83.2

(2) 4h

2

12.8h

x

Breadth = 2 cm

Length = 8 cm

Height = 3.2 cm

Example 3

Solution:

8. Must justify minimum

values using nature table.

X

Slope

( )A x

2

-ve 0 +ve

The surface area is minimised when

the breadth is 2 cm 9. Make statement

(1) 16 128 112A

128(4) 16(4) 56

16A

2

128( ) 16A x x

x

3 8 2x

Example 3

Solution to (c):

10. To find actual maximum area

sub maximum x-value just

found into expression for area.

min 32 64Area (DO NOT USE dy/dx)

2

min

1288(2)

2Area

2

min 96Area cm

2 128( ) 8A x x

x

The surface area is minimised when

the breadth is 2 cm

PROOF

Heinemann , p.112, EX 6R

Q1, 2, 5

Example 4 ( ADDITIONAL - NOT ESSENTIAL)

A channel for carrying cables is being dug out at the side of the road.

A flat section of plastic is bent into the shape of a gutter and placed

Into the channel to protect the cables.

4

0

c

m

100 cm

x

x

The dotted line represents the fold in the plastic, x cm from either end.

(a) Show that the volume of each section of guttering is ( ) 200 (20 )V x x x

(b) Calculate the value of x which gives the maximum volume of gutter

and find this volume.

Example 2 ( ) 200 (20 )V x x x

Solution to part (a):

Require volume so:

V l b h

Length = 100 cm

Breadth = (40 – 2x) cm

Height = x cm 100 (40 2 )V x x

100 (40 2 )V x x

100 2(20 )V x x

200 (20 )V x x

Factorise

bracket

(a) Show that the volume of each section of guttering is

“folded bit” =

as required.

Example 2 ( ) 200 (20 )V x x x

Solution to part (b):

'( ) 4000 400V x x

Maximum and

minimum will

occur at SP’s

(b) Calculate the value of x which gives the maximum

volume of gutter.

( ) 200 (20 )V x x x

2( ) 4000 200V x x x Must prepare for differentiation

At SP’s '( ) 0V x

0 4000 400x

0 400(10 )x

0 (10 )x x = 10 Now prove this is a maximum tp

Must make this statement

Example 2 ( ) 200 (20 )V x x x

Solution to part (b):

(b) Calculate the value of x which gives the maximum

volume of gutter.

Now prove this is a maximum tp x = 10

X 10

0

Slope

dV

dx

10 10

+ve -ve

'( ) 4000 400V x x

So when x = 10 we have a maximum

Example 2 ( ) 200 (20 )V x x x

Solution to part (b):

(b) Calculate the value of x which gives the maximum

volume of gutter.

Now calculate maximum volume ( ) 200 (20 )V x x x

So when x = 10 we have a maximum

(10) 200 10 (20 10)V

3(10) 20000V cm

So maximum volume of guttering

is 20,000 cm3. Make statement